Knowing one probability hands you its opposite for free

0:00The additive axiom is fussy. It only speaks about two events that cannot both occur. For most pairs it has nothing to say at all. But there is one pair it was made for, and every event has one. Take any event, and take everything the experiment can still do instead. Those two share no outcome, because an outcome is either in the event or it is not. And between them they leave nothing out, for the same reason.

0:31That is not an observation about a nice example. On all one thousand and twenty-four events of a ten-outcome list, on all two hundred and fifty-six of an eight-outcome one, and on all sixteen of a four-outcome one, there is not a single exception. Both halves are doing work, and neither implies the other. Two single cards share nothing and cover almost none of the pack. The first seven cards and the last seven cover everything and share four.

1:04Only the pair of an event and what it omits does both at once. Ten cards, numbered one to ten, one drawn. Every card is worth a tenth, and the ten tenths total one. Take the event holding two, four, six and eight. Four cards, each worth a tenth, so the event is worth four tenths. Two fifths. Now write down what it omits. One, three, five, seven, nine, and ten.

1:35Six cards, so six tenths, which is three fifths. Four and six make ten, and two fifths and three fifths make one. Nothing has been proved yet. That is one example, and one example is a coincidence until it is shown not to be. Before the proof, look hard at the six cards that were left. One, three, five, seven, nine. And ten. The ten is even, and it is in there.

2:06The event was never the even numbers. It was four particular even numbers, and there are five available. So a student who names the omitted cards by a property, and says they are the odd ones, is wrong. The odd cards number five, and the omitted cards number six. That answer comes to one half instead of three fifths. It is short by a tenth, and the one card it loses is the ten.

2:35The omitted outcomes are fixed by exclusion. They are whatever is left, and no property of the event's own members can be trusted to name them. How unusual is that? Write down twenty plain descriptions of the ten cards. Even, odd, prime, composite. A perfect square, a perfect cube, a multiple of three, of four, of five. Below five, above five, at most three, at least eight. A single digit, two digits, a factor of ten, a factor of twelve.

3:11A triangular number, a Fibonacci number, and the one card that is neither prime nor composite. Those twenty name twenty different sets of cards. Not one of them names the event we started with. The nearest is even, and it is off by exactly one card, and that card is the ten. Now the sharper count. Of those twenty named sets, only four have an omission that any of the twenty also names.

3:41Sixteen of them leave behind a collection of cards with no name in the list at all. Even the event holding no card is one card short of being nameable. The nearest description to it is two digits, which names the ten alone. Naming is the exception here, not the rule. Now the proof, and it is one line long. Step away from the cards, because the cards are illustration and the proof does not need them.

4:13An event and what it omits cannot both occur. So the additive axiom applies, and their two values add. Between them they hold every outcome, so what they add up to is the value of the sure event. And a second axiom fixes that at one. Therefore the two values total one, and knowing either one hands you the other. Two axioms were used. Additivity across a pair that cannot both occur, and the value of the sure event.

4:45The third axiom, the one saying no value is negative, was not used. Neither was the assumption that the outcomes are equally likely. That is easy to say and easy to doubt, so it is worth measuring. Here is a table on the ten cards that carries a negative entry. One card is worth minus a tenth, and the other nine are worth eleven ninetieths each. It totals one, and it is not admissible, because a probability is never negative.

5:20Run the rule on it anyway. It holds on every one of the one thousand and twenty-four events. So non-negativity really was not what the proof leaned on. Now a table that is perfectly non-negative and totals nine tenths instead of one. The rule fails on every one of the thousand and twenty-four. That is the axiom that was load-bearing. And a table giving the ten cards ten different weights, all legitimate, holds the rule everywhere.

5:54Equal likelihood was never needed. One control, to show the test can fail. Give every event the square of its size over the square of ten. That function is never negative, it gives the sure event one, and it is not additive. The rule fails on one thousand and twenty-two of the thousand and twenty-four events. The two it survives are the event holding nothing and the event holding everything, which is to say the two where there was nothing to add.

6:29So what has been bought? Not a new fact. Every value the rule produces could have been had by totalling the omitted outcomes directly. Those two routes agree on every event of the ten cards, under an even table and under an uneven one. They agree on every event of three tossed coins, and on every event of a four-outcome list weighted a half, a quarter, an eighth and an eighth.

6:59What has been bought is a choice of route. Whenever the omitted outcomes are easier to count, or easier to describe, count those instead and subtract. And notice which of the two things in this video actually needs the equally likely hypothesis. It is not the rule. It is the habit of reading a probability off a count. Size over ten agrees with the even table on all one thousand and twenty-four events, and comes apart from the uneven table on nine hundred and seventy-six of them.

7:37It agrees on only forty-eight. A deck of fifty-two cards, one drawn. Four of them are aces, so an ace is four fifty-seconds, which is one thirteenth. What is not an ace? Subtract, and it is twelve thirteenths. Or total the forty-eight cards that are not aces, and it is twelve thirteenths. The two routes agree, which is the plainest evidence that this is a theorem and not a trick.

8:09But look at the effort. One route counts four cards, the other counts forty-eight, which is twelve times as many. Thirteen cards are diamonds, so a diamond is a quarter and not a diamond is three quarters. Nine discs, four red, three blue, two yellow. Blue is three ninths, which is a third. Not blue is two thirds by subtracting, and two thirds by counting the six discs that are not blue.

8:40Twenty-six cards are black, so a black card is a half. And a card that is not black is also a half. Students read that as an error, and it is not. Of the hundred and one values in hundredths, exactly one equals what remains after it, and it is a half. A number is what is left over from one only at the halfway point. On the ten cards there are two hundred and fifty-two events out of the thousand and twenty-four worth exactly what they omit.

9:15Every single one of them holds five cards. But that is a fact about the even table and not about the number five. Put the uneven weights back on the same ten cards, and not one of the thousand and twenty-four events is worth what it omits. The half is a value, not a size. Now the phrase that always means take the omitted outcomes. At least one. Three coins are tossed, eight outcomes, an eighth each.

9:47How likely is at least one head? Counting directly means one head, or two heads, or three heads. Counting what it omits means no head, which is one outcome, worth an eighth. So at least one head is seven eighths, and the seven outcomes are there to check. At most two tails is the same shape of question. It omits three tails, one outcome, so it is seven eighths as well.

10:15One warning while we are here. At least one head and exactly one head are different events. At least one holds seven outcomes and exactly one holds three, and their values are not equal. Two candidates sit an entrance test. Ana qualifies with probability nought point nought five, Ben nought point one, and both of them nought point nought two. At least one qualifying is the two-event rule, and it is nought point one three.

10:49Now neither of them qualifies. That is what at least one omits, so it is one less nought point one three, which is nought point eight seven. And at least one of them fails. That is what both qualifying omits, so it is one less nought point nought two, which is nought point nine eight. Two complements in one question, taken against two different events, and mixing them up is the commonest mistake here.

11:18The step from neither to the complement of either needs a rule about sets. Neither of two events is the complement of at least one of them. On all four hundred and ninety-six pairs of events on a five-outcome list that holds with no exceptions, and so does its partner. Swapping the two operations round breaks it on every one of the four hundred and ninety-six. Four more, quickly, because the pattern is the whole point.

11:50An event worth two elevenths omits nine elevenths. Two events with a union of five eighths leave neither of them at three eighths. A student on at least one activity is nineteen thirtieths, so a student on neither is eleven thirtieths. And if not both is nought point two five, then both is nought point seven five, which is not nothing, so the two can happen together. Every one of those was reached by totalling the omitted outcomes over their own members, not by subtracting.

12:27You can tell, because the same routine asked about a table totalling nine tenths gives thirty-six hundredths and fifty-four hundredths. Those come to nine tenths, which is exactly what a subtraction could not have produced. Three letters go into three envelopes at random, one each. What is the chance at least one letter reaches the right envelope? There are six arrangements, so write them all out. Sort them by how many letters land right.

13:01Two arrangements put nothing in its own place. Three put exactly one. None put exactly two. And one puts all three. That none is not an accident. If every letter but one is right, the last one has nowhere else to go, so it is right too. For every number of letters from two up to seven, the count of arrangements with exactly one wrong is nought. So nothing in its own place is two arrangements out of six, and at least one right is four out of six, which is two thirds.

13:36Count it directly and you have two cases to add. Count what it omits and you have one. The arrangements with nothing in place go nought, one, two, nine, forty-four, two hundred and sixty-five, one thousand eight hundred and fifty-four. The direct route gets worse quickly, and the other one never does. A hundred students, split into groups of forty and sixty, and two of them are picked. There are four thousand nine hundred and fifty pairs in all.

14:10Pairs from inside the smaller group number seven hundred and eighty. Pairs from inside the larger number one thousand seven hundred and seventy. Together, two thousand five hundred and fifty, so both from the same group is seventeen thirty-thirds. The other question, one from each group, should never be counted directly. It is what the first omits, so it is sixteen thirty-thirds. And it checks out: two thousand four hundred mixed pairs, and the two counts account for every pair there is.

14:47Three last questions, each of which looks like something else. Sixty marbles, ten red, twenty blue, thirty green, and five are drawn. At least one green. Directly, that is five separate cases, one green, or two, or three, or four, or five, and every one of the five terms matters. Drop any single one and the answer changes. What it omits is one case. All five drawn from the thirty that are not green, which is a hundred and seventeen over four thousand four hundred and eighty-four.

15:25So at least one green is four thousand three hundred and sixty-seven over four thousand four hundred and eighty-four. Ten thousand lottery tickets and ten prizes. One ticket wins with probability one in a thousand, so it does not win with probability nine hundred and ninety-nine in a thousand. And a die whose faces carry one, one, two, two, two and three. One face shows a three, so not a three is five sixths.

15:57In each of these you were asked for the bigger side, and the arithmetic counted the smaller one. One last thing, and it is the thing the ten cards were really showing. What an event omits is not a property of the event. It is a property of the event and the experiment together. The same four cards, two, four, six and eight, omit six cards inside a pack of ten.

16:26Inside a pack of twelve they omit eight, and two of those eight are even. The event did not change at all. The experiment did. So do not name what is left by looking at what you took. Name it by looking at what is there. Then subtract, and let the axiom do the rest.